CE258 Chapter 6 Computer Arithmetic

Chapter 6
Computer Arithmetic
CE258: Microprocessor and Computer Organization 1
U & PU Patel Department of Computer Engineering

Contents
• Introduction
• Binary, Octal, Decimal, Hexadecimal representation
• Integer Numbers: Sign-Magnitude
• 1's complement
• 2's complement
• Addition and Subtraction
• Multiplication Algorithm
U & PU Patel Department of Computer
Engineering

Data Types
Data Types:
• Binary information is stored in memory or processor registers
• Registers contain either data or control information
• Data are numbers and other binary-coded information
• Control information is a bit or a group of bits used to specify the sequence of
command signals
• Data types found in the registers of digital computers
• Numbers used in arithmetic computations
• Letters of the alphabet used in data processing
• Other discrete symbols used for specific purpose
• The binary number system is the most natural system to use in a digital computer
• Number Systems
• Base or Radix r system : uses distinct symbols for r digits
• Most common number system :Decimal, Binary, Octal, Hexadecimal
• Positional-value(weight) System : r2 r 1r0.r-1 r-2 r-3
• Multiply each digit by an integer power of r and then form the sum of all
weighted digits
Engineering

Data Types
 Decimal System/Base-10 System
• Composed of 10 symbols or numerals(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
 Binary System/Base-2 System
• Composed of 2 symbols or numerals(0, 1)
• Bit = Binary digit
 Octal System/Base-8 System
• Composed of 8 symbols or numerals(0, 1, 2, 3, 4, 5, 6, 7)
• Bit = Binary digit
 Hexadecimal System/Base-16 System :
• Composed of 16 symbols or numerals(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F)
 Binary-to-Decimal Conversions
1011.1012 = (1 x 23
) + (0 x 22
)+ (1 x 21
) + (1 x 2o
) + (1 x 2-1
) + (0 x 2-2
) + (1 x 2-3
)
= 810+ 0 + 210 + 110 + 0.510 + 0 + 0.12510
= 11.62510
 Octal-to-Decimal Conversions
(736.4)8 = 7 x 82
+ 3 x 81
+ 6 x 80
+ 4 x 8-1
= 7 x 64 + 3 x 8 + 6 x 1 + 4/8 = (478.5)10
 Hexadecimal-to-Decimal Conversions
(F3)16 = F x 16 + 3 = 15 x 16 + 3 = (243)10
Engineering

Data Types
 Conversion of decimal 41.6875 into binary
Repeated division
Integer = 41
41 / 2 = 20 remainder 1
20 / 2 = 10 remainder 0
10 / 2 = 5 remainder 0
Read the result upward to give an answer of (41)10 = (101001)2
Fraction = 0.6875
0.6875 x 2 = 1.3750 integer 1
1.3750 x 2 = 0.7500 integer 0
0.7500 x 2 = 1.5000 integer 1
1.5000 x 2 = 1.0000 integer 1
Read the result downward (0.6875)10 = (0.1011)2
(41.6875)10 = (101001.1011)2
Engineering
(ignore the integral part and multiply only the
fractional part until you get 0’s in fractional
part )
(binary number will end with 1) : LSB
(binary number will start with 1) : MSB
MSB
LSB

Data Types
Hex-to-Decimal Conversion
2AF16 = (2 x 162
) + (10 x 161
) + (15 x 16o
)
= 51210 + 16010 + 1510
= 68710
Decimal-to-Hex Conversion
42310 / 16 = 26 remainder 7 (Hex number will end with 7) : LSB
2610 / 16 = 1 remainder 10
110 / 16 = 0 remainder 1 (Hex number will start with 1) : MSB
Read the result upward to give an answer of 42310 = 1A716
Hex-to-Binary Conversion
9F216 = 9 F 2
= 1001 1111 0010
= 1001111100102
Binary, octal, and hexadecimal Conversion
Engineering
 Binary-to-Hex Conversion
1 1 1 0 1 0 0 1 1 02 = 0 0 1 1 1 0 1 0 0 1 1 0
3 A 6
= 3A616
Hex Binary Decimal
0 0000 0
1 0001 1
2 0010 2
3 0011 3
4 0100 4
5 0101 5
6 0110 6
7 0111 7
8 1000 8
9 1001 9
A 1010 10
B 1011 11
C 1100 12
D 1101 13
E 1110 14
F 1111 15
14 0001 0100 20
F8 1111 1000 248

Data Types
• Binary-Coded-Decimal Code
• Each digit of a decimal number is represented by its binary equivalent
8 7 4 (Decimal)
1000 0111 0100 (BCD)
• Only the four bit binary numbers from 0000 through 1001 are used
• Comparison of BCD and Binary
13710 = 100010012 (Binary) - require 8 bits
13710 = 0001 0011 0111BCD (BCD) - require 12 bits
• Alphanumeric Representation
• Alphanumeric character set
• 10 decimal digits, 26 letters, special character($, +, =,….)
• ASCII(American Standard Code for Information Interchange)
• Standard alphanumeric binary code uses seven bits to code 128 characters
Engineering

ASCII Table
Engineering
Character Binary code Character Binary code Character Binary code Character Binary code
A 100 0001 U 101 0101 0 011 0000 / 010 1111
B 100 0010 V 101 0110 1 011 0001 , 010 1100
C 100 0011 W 101 0111 2 011 0010 = 011 1101
D 100 0100 X 101 1000 3 011 0011
E 100 0101 Z 101 1010 4 011 0100
F 100 0110 5 011 0101
G 100 0111 6 011 0110
H 100 1000 7 011 0111
I 100 1001 8 011 1000
J 100 1010 9 011 1001
K 100 1011
L 100 1100
M 100 1101 space 010 0000
N 100 1110 . 010 1110
O 100 1111 ( 010 1000
P 101 0000 + 010 1011
Q 101 0001 $ 010 0100
R 101 0010 * 010 1010
S 101 0011 ) 010 1001
T 101 0100 - 010 1101

Complements
• Complements are used in digital computers for simplifying the
subtraction operation and for logical manipulation
• There are two types of complements for base r system
• 1) r’s complement 2) (r-1)’s complement
• Binary number : 2’s or 1’s complement
• Decimal number : 10’s or 9’s complement
• (r-1)’s Complement
• (r-1)’s Complement of N = (rn-1)-N
• 9’s complement of N=546700
(106-1)-546700= (1000000-1)-546700= 999999-546700
= 453299
• 1’s complement of N=101101
(26-1)-101101= (1000000-1)-101101= 111111-101101
= 010010
• r’s Complement
• r’s Complement of N = rn-N for N != 0 and 0 for N=0
• 10’s complement of 2389= (104 -2389)= (10000-2389) =7611
• 2’s complement of 1101100= 0010011+1= 0010100
Engineering
N : given number
r : base
n : no of digits in the
given number
546700(N) + 453299(9’s com)
=999999
* r’s Complement
(r-1)’s Complement +1 =(rn-1)-N+1= rn-N
101101(N) + 010010(1’s com)
=111111

Subtraction of Unsigned Numbers
• Subtraction of Unsigned Numbers
• 1) M + (rn-N)
• 2) M  N : Discard end carry, Result = M-N
• 3) M  N : No end carry, Result = - r’s complement of (N-M)
• Decimal Example
72532(M) - 13250(N) = 59282
72532
+ 86750 (10’s complement of 13250)
1 59282
Result = 59282
• Binary Example
1010100(X) - 1000011(Y) = 0010001
1010100
+ 0111101 (2’s complement of 1000011)
1 0010001
Result = 0010001
Engineering
M  N
Discard
End Carry
X  Y
13250(M) - 72532(N) = -59282
13250
+ 27468 (10’s complement of 72532)
0 40718
Result = -(10’s complement of 40718)
= -(59281+1) = -59282
M  N
No End Carry
1000011(X) - 1010100(Y) = -0010001
1000011
+ 0101100 (2’s complement of 1010100)
0 1101111
Result = -(2’s complement of 1101111)
= -(0010000+1) = -0010001
X  Y
Discard
End Carry
No End
Carry

Practice Examples
1. Convert the following binary numbers to decimal :
a. 101110 b. 1110101 c. 110110100
2. Convert the following numbers with the indicated bases to decimal :
a. (12121)3 b. (4310)5 c. (50)7
3. Obtain 9’s & 10’s complement of the following eight-digit decimal numbers:
a. 12349876 b. 00980100 c. 90009951
4. Perform the subtraction with the following unsigned decimal numbers by taking the 10’s complement of
the subtrahend
a. 5250-1321
b. 1753-8640
c. 20-100
d. 1200-250
5. Perform the subtraction with the following unsigned binary numbers by taking the 2’s complement of the
subtrahend
a. 11010-10000
b. 11010-1101
c. 100-110000
d. 1010100-1010100
Engineering

Thank You.
Engineering

Addition ad Subtraction
• All the arithmetic operations can be performed by
addition,subtraction,multiplication and division
• Addition and Subtraction operations are performed on
• Signed magnitude data: 1 bit is used to represent the sign and other
bits represent the magnitude.
• 2’complement data
Engineering

Addition and Subtraction with Signed-Magnitude Data
• Addition Algorithm:
When the signs of A and B are identical ,add the 2 magnitudes and attach the sign of A to the result
When the signs of A and B are different, compare the magnitudes and subtract the smaller number from larger.
• Subtraction Algorithm:
When the signs of A and B are different ,add the 2 magnitudes and attach the sign of A to the result
When the signs of A and B are identical, compare the magnitudes and subtract the smaller number from larger.
• Choose the sign of the result to be same as A if A>B or complement of A if A<B
Engineering
Operation Add
Magnitudes
Subtract Magnitudes
When A>B When A<B When A=B
(+A) + (+B) +(A+B)
(+A) + (-B) +(A-B) -(B-A) +(A-B)
(-A) + (+B) -(A-B) +(B-A) +(A-B)
(-A) + (-B) -(A+B)
(+A) - (+B) +(A-B) -(B-A) +(A-B)
(+A) - (-B) +(A+B)
(-A) - (+B) -(A+B)
(-A) - (-B) -(A-B) +(B-A) +(A-B)

Hardware implementation
A and B be two registers that hold the
magnitudes of No
As and Bs be two flip-flops that hold the
corresponding signs
The Result is transferred into A and As.
Parallel adder is needed to perform the micro
operation A + B.
Output carry are transferred to E flip-flop
• Where it can be checked to determine the
relative magnitude of the Nos.
Add overflow flip-flop (AVF) holds the overflow
bit when A and B are added.
Engineering
B register
Complementer
Parallel Adder
A register
Bs
As
E
AVF
Load Sum
M
(mode
control
)
Output carry
S
M=0,Input Carry=0  A+B
M=1,Input Carry=1  A+B’+1
Input carry

Hardware Algorithm
Engineering
Addition:
Signs are same ->add
Signs are different->subtract
Subtraction:
Signs are same ->subtract
Signs are different->add

Addition and Subtraction
• 2’s Complement data
• Left most bit of a binary number represents sign-bit. (0 for =ve
and 1 for –ve)
• If sign bit=1, then the entire number is represented in
2’complement form.
• Example +33 00100001
-33 11011111
• Addition:
• Add the numbers and treat the sign bits same as other bits
• Carry out of the sign bit position is discarded
• Subtraction:
• Take the 2’complement of the subtrahend and add it to minuend
Engineering

Addition and Subtraction
Engineering
• Overflow: When two n-bit numbers are added and the sum
occupies n+1 bits
• The overflow can be detected by applying the last two carries
out of the addition to an XOR gate. A ‘1’ at the output indicates
an overflow.

Hardware and Algorithm for Signed 2’s complement Data
Engineering

Engineering

Multiplication Algorithm
Engineering

Hardware for Multiply operation
Engineering

Flow Chart for Multiply operation
Engineering

Booth Multiplication Algorithm
• Booth algorithm gives a procedure for multiplying binary
integers in signed 2’s complement representation in efficient
way, i.e., less number of additions/subtractions required.
• It operates on the fact that strings of 0’s in the multiplier require
no addition but just shifting and a string of 1’s in the multiplier
from bit weight 2^k to weight 2^m can be treated as 2^(k+1 ) to
2^m.
• Example: (+14) is represented as 001110 has string of 1’s from
23 to 21.
Here K=3,m=1
(+14) can be represented as 2k+1-2m=24-21=16-2=14.
MX14 = MX24-MX21
Engineering

Booth Multiplication Algorithm
• As in all multiplication schemes, booth algorithm requires
examination of the multiplier bits and shifting of the partial
product.
• Prior to the shifting, the multiplicand may be added to the partial
product, subtracted from the partial product, or left unchanged
according to following rules:
• The multiplicand is subtracted from the partial product upon
encountering the first least significant 1 in a string of 1’s in the multiplier
• The multiplicand is added to the partial product upon encountering the
first 0 (provided that there was a previous ‘1’) in a string of 0’s in the
multiplier.
• The partial product does not change when the multiplier bit is identical
to the previous multiplier bit.
Engineering

Hardware Implementation of Booths Algorithm
Engineering
We name the register as A, B and Q, AC, BR
and QR respectively.
Qn designates the least significant bit of
multiplier in the register QR.
An extra flip-flop Qn+1is appended to QR to
facilitate a double inspection of the multiplier.

Flowchart
Engineering

Practice Examples
• Show the contents of E,A,Q and SC during the process of
multiplication of two binary numbers, 11111 (multiplicand) and
10101(multiplier). The signs are not included.
• Show the step-by-step multiplication process using Booth
algorithm when the following binary numbers are multiplied.
Assume 5-bit registers that hold signed numbers.
a. (+15) X (+13)
b. (+15) X (-13)
Engineering

CE258 Chapter 6 Computer Arithmetic

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